Your paragraph text
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Chapter 2
2.1 (a)
k1 0 –k1 0 0000
(1)
[k ] =
–k1 0 k1 0 0 0 0 0 0 0 0 0 0 0 0 0 (2) [k ] = 0 0 k2 –k2 0 0 –k2 k2 0 0 0 0 (3)
[k 3 ] =
0 k3 0 – k3 0 0 0 0 0 –k3 0 k3
(3) (1) (2) [K] = [k ] + [k ] + [k ]
[K] =
0 – k3 – k2 k 2 k3
k1 0
– 0 k1 k 3 0 k10 k2 –k1 0 –k3 – k2
(b) Nodes 1 and 2 are fixed so u1 = 0 and u2 = 0 and [K] becomes
k k–
[K] = 12
k2 – k 2 k 2 k3
{F} = [K] {d}
k =1 –
Fx 3 x F 0 k1= 23 2– 4 P – 2 k2 3 4
⇒
{F} = [K] {d}
k2
k2
k2 –k2 k3
k
k
k
u 3 u u 4 u
k
⇒ [K–1] {F} = [K–1 ] [K] {d}
⇒ [K–1 ] {F} = {d}
Using the adjoint method to find [K–1 ] C + = (– 1)311 =k2k3C21 (– k2) C= (– 1)1 + 212 (– k2) = k2 C22 = k1 + k2 3 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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k2 k1 k
k2 k3 k2 k2 k1k 2 2 det [K ] = | [K] | = (k1 + k2) (k2 +k3) – ( –k 2) (– k2) | [K] | = (k1 + k2) (k2+ k3) – k22 k k3 [C] = 2 k
⇒
T
and C =
2
T
[K–1 ] = [ C ] det K
k2k3 k2 k2 k3 k 2 k2 k1k2 k2 k1 k –1 2 [K ] = = 2 k2)(k2 k3) – k k1 k2 k1 k3 k k (k1 2 3 2
k2 0 u k1k k2 2 P = 3 k1 k2 k1 k3 k 2k3 u k2P 4 u3 = k1k2 k1k k2 k3 3 k k P ( u 12)4 = k k 2k k 3 k k 1 2 1 3 (c) In order to find the reaction forces we go back to the global matrix F = [K] {d} 2k
k3
⇒ ⇒
k1
Fx 1x
=
Fx 2x
0 k3
0 k1
0 k3
0
F
k1 0 k1 k 2 k2
0 k3 k2 k2
u 1 u k
3
2
u k2P k1k2 k1 k3 k k 3 2 3 u k1 k2 P 1
3 F1x = – k1 u3 = – k F
⇒4 F x = k1k2 k1 k3 k2 k3
4
1
(k F2x = – k3 u4 = – k3 1 k k1k2 k1k
k2) P ⇒ F x = k1k k (k1 k1 k3 k2 k3 2
2 )P 3 k
2
k3
3
2
2.2
k1 = k2 = k3 = 1000
lb in.
(1) (2) (2) (3)
(1) k k (1) (2) k k (2) [k] = ; [k] = (2) kk kk (3) By the method of superposition the global stiffness matrix is constructed. 4 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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(1) (2) [K] =
k k 0
Node 1 is fixed {F} = [K] {d}
1x 0 Fx ? 2
k k k k
0(1) k(2) k(3)
⇒
0 k
u
0
1
?
k
u
⇒ u1 = 0 and u3 = δ k
?
Fx
(3)
=
k 0
0
⇒F F3x = 2 kk ⇒3 u k
k 2k k
k k
0
20
u2
2ku 2 ku
u F3x 3
⇒
1in. 2 = = = u2 = 0.5″ 2k2 2 F x = – k (0.5″) + k (1)″
k k
[K] =
k 2k k
0 k k
k k 2
lblb (– 1000 ) (0.5) +″ (1000 ) (1″) 3 xin.=in.
= 500 lbs F3x Internal forces 3 Element (1) F f1x (1)
k u 0 1 0.5 u lb f (1)1x = (– 1000 )2(0.5) ″ in.
f2x(2)
⇒
=
k k
f (1)2x= (1000
k
lb ″ ) (0.5) in.
⇒f
⇒ fx
1x
2
(1)
(1)
= – 500 lb
= 500 lb
Element (2)
f 2x(2) f 3x
(2)
=
k k
2.3
k u k 2 u 3
0.5 1
⇒ ff
2x 3x
(2)
– 500 lb
(2)
500 lb
(1) (2)
= [(3)] = [(4) k k (a) [k] = [k]kk] = k k By the method of superposition we construct the global [K] and knowing {F} = K [ ] {d} we have Fx
?
1x
0
Fx 2x
P = 0
Fx
?
k k 0 0 0
k 2k k 0 0
0 k 2k k 0
0 0
0 0
u
k 2k k
0 k
u
3 F
5
k
0
1 2 u 0 3 u
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4
4
F
u
5
5
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(b)
0 u 2ku2 ku3 k 2 P ku2 2ku3 ku4 u ku3 2ku4 0 3 2k u u3u32 = ; u4 = 2 2 u 4 in the middle Substituting in the equation
0 2k P = k 0 0
⇒
k 2k k
0
P = – k u2 + 2k u3 – k u4
u u P = – k 3 + 2k u – 3 2 3k 2
⇒ P = k u3 ⇒ u3 = ⇒
P k
P P k k (c) In order to find the reactions at the fixed nodes 1 and 5 we go back to the global equation {F} = [K] {d} P P k F1x = – ku2 = – F1x = k 2 2 P ku F5 P x= F5x = – 4 = –k 2 2k Check u2 = ; u4 = 2 2
⇒ ⇒
ΣFx = 0
⇒ F1x + F5x +P = 0 ⇒ P P +P=0 + ⇒ 2 2 0=0
2.4
(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =
k k
k k
By the method of superposition the global [K] is constructed.
Also {F} = [K] {d} and u1 = 0 and u5 = δ 0 u 1 ? k 0 Fx = 0 u ? 0 2 ? 2k k 0 0 2x 0 u k 0 0 Fx k 3 3 ? u F 6 4 © 2012 Cengage Learning.4 All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. u F 5 5
Fx
? 1x 0
k k
k 2k k
0 k 2k k 0
0 0
0 0
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(1)
(b) 0 = 2k u2 – k u3 0 = – ku2 + 2k u3 – k u4
(2)
0 = – k u3 + 2k u4 – k δ
(3)
From (2) u3 = 2 u2 From (3)
2u u = 24
2
Substituting in Equation (2)
2 x ⇒– k (u2) + 2k (2 u2) – k 2 ⇒ – u2 + 4 u2 – u2 – = 0 ⇒ u2 = 2
4
2
⇒ u 3 =2 ⇒ u3 = 4 2 ⇒ 2 4 ⇒
3 4
u4 = u4 = 2
(c) Going back to the global equation {F} = [K]{d}
F1x = – k u2 = k
4
⇒ F1x
F5x = – k u4 + k δ = – k
3 4
=
k 4 +kδ
⇒ F x = k4 5
2.5 2 1
1
2
kip 1
(1)
[k ] =
(3)
[k ] =
(5)
[k ] =
kip in. 2 3
4 kip 4 in.
in.
5
4 5
d1 d2 1 1 ; [k 1 1
d2 d4 2 2 (2) ]= 2 2
d2 d4 3 3 ; [k 3 3
d2 d 4 4 4 4 4
d 4 5 5
(4)
]=
kip in.
3
kip
3
x
in.
d3 5
5 7
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Assembling global [K ] using direct stiffness method
[K] =
1 1 0 0
1 1 2 3 4 0 2 3 4
1 1 0 0
1 10 0 9
0 2 3 4
0 0 5 5
5 234 5
Simplifying
[K] =
0 0 5 5
0 9kip 5in. 14
2.6 Now apply + 2 kip at node 2 in spring assemblage of P 2.5.
∴ F2x = 2 kip [K]{d} = {F}
[K] from P 2.5
1 10 0 9
u 0 0 F1 9 1 = 2 5 u 0 F3 1 2 0 4 u where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed. 3 of (A) Using Equations (1) and (3) u 2 u 10 9 4 = 9 14 2 0 1 1 0 0
Solving
0 0 5 5
u 4
u2 = 0.475 in.,
(A)
u4 = 0.305 in.
2.7
conv.
f1x = C, f2x = – C f= – kδ = – k(u2 – u1)
∴
f1x = – k(u2 – u1) f2x = – (– k) (u2 – u1)
f2x
k –k
[K] =
k –k
f1x =
∴
u 1 u same as for 2 –k tensile element k – k k
8 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2.8 k = 500in. lb
k = 500in. lb
k1 = 500
1 1
1 ; k2 = 500 1
[K] = 500
1 1 0
1 2 1
500 lb
1 1
1 1
So
0 1 1
{F} = [K] {d} 1
⇒ 3
F 2F F
⇒
? 0 = 500 1000
0 u 1 ? u ? 2 u 3
1 1 0 1 1 2 0 1 1
0 = 1000 u2 – 500 u3 500 = – 500 u2 + 500 u3 From (1)
500 1000
u2 = u3
(1) (2)
⇒ u2 = 0.5 u3
(3)
Substituting (3) into (2)
⇒ 500 = – 500 (0.5 u3) + 500 u3 ⇒ 500 = 250 u3 ⇒ u3 = 2 in. ⇒ u2 = (0.5) (2 in.) ⇒ u2 = 1 in. Element 1–2
f1x (1) f2x (1)
= 500
1 1
Element 2–3
f 2x(2) f 3x (2)
1 1
= 500
F1x = 500 [1 – 1 0]
in . in . 1 1in. 1 2in.
f
1 0 1 1
f
f f
0 1in. F 2in. 1x
(1)
500 lb
(1)
500 lb
(2)
500 lb
(2)
500 lb
1x 2x
2x 3x
500lb
2.9 lb in.
lb in.
lb in.
9 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(1)
[k ] =
(2)
[k
]=
(3)
[k ] =
[K] =
Fx
?
1x
1000
Fx
0
2x
4000
F 3
⇒
=
(1) 1000 1000
(2) 1000 1000
(2) 1000 1000 (3) 1000 1000
(3) 1000 1000 (4) 1000 1000
(1) 1000 1000 0 0
(2) 1000 2000 1000 0
(3) 0 1000 2000 1000
(4) 0 0 1000 1000
1000 1000
1000 2000
0 1000
0 0
u 0
0 0
1000 0
2000 1000
1000 1000
u
4 Reactions
Element forces Element (1)
f1x (1) f2x (1) Element (2) f 2x(2) f 3x (2) Element (3) f 3x (3)
2.10
f
4x
(3)
=
=
=
100 0 100 0 100 0 100 0 100 0 100 0
2 u
u = 0 in. 1 = 3 in. = u 7 in. 2 = 11 in. u 3 u F41x = [1000 – 1000 0 0]
F
1
3 u 4
u 0 1 3 u 7 2 1 u 1 3 1 000 0 u f1x (1) 1000 3 4 f (1) 2x
⇒
1 000 3 1000 7
⇒ ff
1000 1000
⇒ ff
7 11
2x 3x
⇒ F1x = – 3000 lb
3000 lb 3000 lb
(2) (2)
3x 4x
4000lb 4000 lb
(3) (3)
4000 lb 4000 lb lb in.
lb in.
lb in. 10 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(1)
[k ] = (2)
[k ] =
100 0 100 0
500 50 [k(3)] = 500 0 {F} = [ K ]50{d} 0 ? 1000 – 4000 1000 = ? 0 ? 0
Fx 1x Fx 2 x F 3 F Reactions 4
⇒
u2 =
Fx 1x Fx
⇒
Element (1) F 3 f1x (1) F f2x4(1)
f 3x
(2)
f 4x 2.11
(3)
500 500
1000 2000 500 500
=
2000 4000 1000 1000
lb
=
=
=
0 500 500 0
100 0 100 0 50 0 50 0 50 0 50 0
1000 1000
0
u
0 500 0 500
1
?
u
0
2 0 u
4000 = – 2 in. 2000
=
Element (3)
f 2x (3)
500
1000 2000 500 500
Element (2)
f 2x(2)
500
1000 1000 0 0
2x F Fx 3 1x F Fx 4 2x
1000 1000
3 u 4 0 0 2 500 0 0 0 500
0 500 500 0
0 –2
⇒ ff
1x 2x
2 500 500 0
⇒ ff
2 500 500 0
⇒ ff
N m
N m
2x 3x
2x 4x
(1)
=
(1)
(2) (2)
(3) (3)
=
2000 lb 2000
–1000 lb 1000
1000 lb 1000
d = 20 mm
11 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20 00 (2) ; [k ] = 2000
200 0 {F} = [ K ]200 {d} 0 Fx ? 2000 2000 1x 0 = 0 Fx ? [k(1)] =
⇒
2
u2 = 0.01 m
F Reactions 3
F1x= (– 2000) (0.01)
Element (1)
f
1x
ˆ Element (2)
f
2x
ˆ f
2x
ˆ f
2.12
=
3x
2000 2000
u 1 u 2 u 3 F1x = – 20 N
2000 0 4000 20 00 2000 2000
⇒
2000 2000
0 0.01
20 00 200 2000 0 200 0 0 ? 0 . 0 2 m
f
⇒ ˆ f
=
2000 –2 00 –2 00 0 0 2000
0.01 0.02
1x
=
2x
ˆ f
⇒ ˆ
2x
f
ˆ
=
3x
ˆ N m 2
[k(2)] = 10000
2 2
2
1 1 0 0
1 3 2 0
N
3
1
1 1
(1) (3) [k ] = [k ] = 10000
N
N m
N m
1
2 0 2 0 2 0 2 0
1
2
{F} = [K] {d}
Fx 1x Fx 2 x F 3 F 4
? 4500 N = 10000 0 ?
0 = – 2 u2 + 3 u3
0 0 2 0 3 1 1 1
⇒ u2 =32u3 ⇒
450 N = 30000 (1.5 u3) – 20000 u3
⇒N 450 N = (25000 ) u ⇒ = 1.8 × 10–2 3u3m m ⇒ u = 1.5 (1.8 × 10–22) ⇒ u2 = 2.7 × 10–2 m
0
u 1 ? u ?
2 0 u u23= 1.5 u3 u 4
Element (1) = 10000 2
f
1x
ˆ f
2x
ˆ
1 1
1 1
0 2.7 10
ˆ
⇒ ˆff
1x 2x
(1) (1)
270N 270N
12
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Element (2)
f
2x
ˆ
= 20000
3x
Element (3) f ˆf
3x
ˆ f
Reactions
= 10000
4x
1 1
1 1
1 1
2.7 10
f
2
⇒ˆ
1.8 10 2
1 1
f
f
ˆ 0 {F1x} = (10000 N ) [1 – 1] m 2.7 10 2
2.13
⇒ F4x = – 180 N kN m
kN m
3x 4x
[k(1)] = k[(2)] = [k(3)] = [k(4)] = 20
180N
(2)
180N (3)
180N
(3)
180N
ˆ
⇒ F x= – 270 N 1
kN m
5 kN
(2)
2
1.8 10 0
N {F 4x} = (10000 ) [–1 1] m
3x
fˆ
⇒ˆ
2
1.8 10 0
2x
kN m
1 1
1 1
{F} = [ K] {}d
Fx ? 1 1 0 0 0 1x 0 1 2 1 0 0 F x 10kN = 20 0 1 2 1 0 2 0 0 0 1 2 1 x F ? 0 0 0 1 x 1 3 0 2u – u u 0.5 u F = u4 u 0 –u 2 u2323 u u20.5 3 4 44 3 F 5 kN = – 20 u2 + 40 (2 u2) – 20 u2 5 5 = 40 u2 u2 = 0.125 m
0 u ? 1 u ? 2 ? u 0 3 u 4 u 5
⇒
⇒
⇒ ⇒ ⇒ u4 = 0.125 m ⇒ u3 = 2(0.125) ⇒ u3 = 0.25 m Element (1)
f
1x
ˆ Element (2)
f
= 20
2x
ˆ f
2x
ˆ f
3x
ˆ
1 1
1 1
0 0.125
⇒ fˆ
1x 2x
(1) (1)
2.5 kN 2.5 kN
f
= 20
1 1
1 1
0.125 0.25
ˆ f
⇒ˆ f
13
2x 3x
(2) (2)
2.5 kN 2.5 kN
ˆ
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Element (3)
f3x = 20 f4x
1 1
1 1
0.25 0.12 5
⇒ ff
f
1 1
1
0.125
⇒ ffˆ
Element (4) 4x
= 20
ˆ f
5x
ˆ
1 0
2.5 kN
(3)
ˆ
4x
(4)
2.5kN 2.5kN
(4)
2.5kN
⇒ F1x = – 2.5 kN
0.125
⇒ F5x = – 2.5 kN
0.125
F5x = 20 [–1 1]
4x
(3)
5x
0
F1x = 20 [1 –1]
3x
0
2.14 N m
4000 NNm/m
1000 N
1 1
[k(1)] = k[(2)] = 400
2000 N
1 1
{F} = [ K]d}{
0 u 1 ? u ? 2 u 3
Fx ? 1 1 0 1 1 x 100 = 400 1 2 Fx 0 1 200 1 2 100 = 800 u2 – 400 u3 F – 2003= – 400 u2 + 400 u3 – 100 = 400 u2 u2 = – 0.25 m
⇒
= 400
f
1 1
1x
ˆ
Element (2) f f̂
2x
2x
ˆ Reaction
⇒
100 = 800 (– 0.25) – 400 u3 u3 = – 0.75 m
Element (1)
f
3x
= 400
1 1
0 0.25
1 – –1 1 1
0.25 0.75
⇒
fˆ (1) 100N 1x
ˆ (1) f2x f
⇒ˆ f
ˆ {F1x} = 400 [1 –1]
0 0.25
2x 3x
100N
(2)
200N
(2)
200N
⇒ F1x ˆ= 100 N
2.15 kN
500 m
kN
1000 m kN 500 m
14 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
(1) [k ] =
500 (2) ; [k ] = 500
50 0 50 500 00
500 (3) ; [k ] = 500
50 0 050 00
u Fx ? 0 500 0 1 0 1x ? 500 500 = u F x 2kN 500 500 2000 1000 ? 2 2 ? 0 0 1000 1000 x u 0 F u3 = 0.001 m 3 3 Reactions u F F1 = (– 500) (0.001) F1x = – 0.5 kN 4 4 x = (– 500) (0.001) = F2x = – 0.5 kN x = – 1.0 kN F4 F2 (– 1000) (0.001) Element (1) x f f F4 50 50 0 0.5 kN 1x 1x = = ˆ ˆx 0 0 0 .0 01 0.5 kN
100 0 100 0
10 00 1000
⇒
⇒ ⇒ ⇒
Element (2) f
⇒
3x
fˆ 2x
=
ˆ Element (3) f f̂
3x
3x
=
ˆ 2.16
f
4x
50 050
500
0 50 0 1000 1000
f 0 0 .0 01
3x
⇒ ˆfˆ
2x
f 1000 1000
0.001 0
⇒ ˆ fˆ
3x
F 1x 100 = 100 F4 x
100 100
=
4x
1kN 1kN
–100 lb
2
lb
=
ˆ
100 lb 100 in.
0.5 kN 0.5 kN
=
3x
f
ˆ 1
50 0 500
3
lb
100 in.
100 100 0 0
100 100 100 100 0
200 100
100 200
1 u2 = in. 3 1 u3 = – in. 3
0 100 100 100 100
4
lb
100 in.
0 0 100 100
u 2 u 3
0 u 2 u0 3
2.17 300
400
500 300 1
1000 N 3
400 4
2 15
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
F1x ? 0 1000
=
400 300
500
500 300 –300–300 –400
0
N F4x?
–500
500
0
0
0
–300–300
–400
(300 300 400)
–400
–400
400 400
0 = 1500 u2 – 600 u3
u 0 1 u 2 0 u 3
1000 = – 600 u2 + 1000 u3
u
1500 u3 = u2 = 2.5 u2 600
4
1000 = – 600 u2 + 1000 (2.5 u2) 1000 = 1900 u2
1000 u2 1= = mm = 0.526 mm 1900 1.9 1 1.9
u 3 = 2.5
F1x = – 500
mm = 1.316 mm
1 1.9
= – 263.16 N
1 F4x= – 400
1.9 – 400
2.5
1 1.9
1 2.5 = – 400 = –736.84 N 1. 1.9 9 ΣFx = – 263.16 + 1000 – 736.84 = 0 2.18 (a) 1000 lb
lb
2000 in.
x
As in Example 2.4 πp = U + Ω 1 k x2, Ω = – Fx U= 2 Set up table 1 (2000 x– 1000 x π )2 p= 2
= 1000 x2 – 1000 x
Deformation x, in.
πp, lb⋅in. 6000 3000
– 3.0 – 2.0 16
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
– 1.0 0.0 0.5 1.0 2.0 p
x
1000 0 – 125 0 1000
= 2000 x– 1000 = 0
⇒ x = 0.5 in. yields minimum πp as table verifies. pp, lb.in.
–3
–2
x, in. 1 2
–1
3
M i n i m u m
(b) xx lb
500 in .
πp =
1000 lb
1 2 2 kx – Fx = 250 x – 1000 x 2
x, in. πp, lb⋅in. – 3.0 11250 – 2.0 3000 – 1.0 1250 00 1.0 – 750 2.0 – 1000 3.0 – 750 p = 500 x – 1000 = 0
⇒
x
= 2.0 in. yields πp minimum x
(c)
N mm
x
– 400 kg ¥ 9.81 m s2= 3924 N
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πp =
1 2 (2000) x2– 3924 x = 1000 x – 3924 x 2
p
⇒
= 2000 x – 3924 = 0 x
x = 1.962 mm yields πp minimum
1 πp min 2
= (2000) (1.962)2 – 3924 (1.962)
⇒ πp =min– 3849.45 N⋅mm 1 x2 – 981 x π (400) p= 2
(d)
p
⇒
= 400 x – 981 = 0
x
x= 2.4525 mm yields πp minimum
1 (400) (2.4525)2 – 981 (2.4525) πpmin= 2
⇒ πp = – 1202.95 N⋅mm min
2.19 F = 1000 lb
x
Now let positive x be upward k = 500 lb in.
π p= π
1 = (500) x2 p– 1000 x 2
π
p = 250 x2 – 1000 x p
x
1 2 kx – Fx 2
= 500 x – 1000 = 0
⇒ x = 2.0 in. ↑ 2.20 1000 lb in. F = 500 lb
F = kδ2
(x = δ)
dU= F dx 18 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
x
U=
0
F
2
(kx) dx 4000
kx3
U=
1000
3
Ω = – Fx
πp = p
x
F = Kd
1
2
2
d
1 3 kx – 500 x 3 2
= 0 = kx – 500
0 = 1000 x2 – 500
⇒
x = 0.707 in. (equilibrium value of displacement)
1 (1000) (0.707)3 –500 (0.707) 3
πp min =
π lb⋅in. p min = – 235.7 2.21 Solve Problem 2.10 using P.E. approach
lb
500 in. lb
1000 in.
4000 lb
lb
500 in.
πp =
3
1 1 1 (e) ) πp = 2 k1(u2 –u2)+1 2 k2(u3 – u2)+2 2 k3(u4 – u22 e 1 – f1x(1)u 1 –f2x (1) u2 –f2x (2) u2
– f3x p
u 1
u
p 2
p
u3 p
u4
(2)
u3 –f2x
(3)
u2 –f4x (1)
= – k1 u2 +k1 u1 – f1x
(3) 4
u
=0
(1)
=ku2 1 – k1 u1 – 2k 3u+ 2 xk u2– k3 u4 +k3 u2 – f2x(1)–
f (2) 2–
f (3) 2x = 0
(2)
(2)
=0
(3)
(3)
=0
(4)
= k2 u3 – k2 u2 – f3x = k3 u4 – k3 u2 – f4x
19 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
In matrix form (1) through (4) become
k1 k1
k
1
0 k2
k1 k 2 k 3 k 2 k
0 0
0
0 k 3 0 k3
0 500 500 0
0 –500 0 500
k2
3
f 1x (1) u f (1)2x f 2x(2) 1 = u f (2) 3x 2 f (3) 4x u 3 u 0 F 1x u 1 4000 4 u 0 = Fx 2 0 3 x F u 4 3
or using numerical values
1000 1000 0 0
1000 2000 500 500
Solution now follows as in Problem 2.10
f (3)2x
(5)
(6)
Solve 2nd of Equations (6) for u2 = – 2 in. For reactions and element forces, see solutionuto Problem 2.10 4 Solve Problem 2.15 by P.E. approach 2.22 kN
500 m
kN
1000 m kN
500 m
πp =
3 e 1
πp
(e)
=
1 1 k 2(u– u232) k1(u3 –u2)+1 2 2 1 + k22 3 (u4 – u3 )–
(1)
f1x
u1
–f(1)(2)3x u3 – f2xu 2 – f3x(2) u3 4 –f(3)(4)3x u3 – f3xu
=0=–
k1 u3 +k1 u(1)1 – f1x
p
u1 p
u2
(2)
p
= 0 = k1 u3 +k2 u3 –k2 u2 –k3 u4 +k3 u3 – f3x – f3x
p
= 0 = k3 u4 –k3 u3 – f3x
u3 u
= 0 = – k2 u3 +k2 u2 – f2x
(2)
(3)
(1)
– f3x – k1 u1
(4)
4
In matrix form
k1
k
u F1x 0 k2 1 k F2x = 2 k 1 k2 k u F3x k 2kN 3 3 1 k2 k 2 F4x k3 0 0 k 3 u For rest of solution, see solutions of Problem 2.15. 3 20 u © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 4 0
1
0 0
2.23 I =a +a2 x I(0) = 1 = I1 I()L= a2
a + a2L =I
2
=1 I2 I 1 a L
∴ I2 I = 11+
II1 L x
Now V = IR V = – V1 = R (I2 – I1) V = V2 = R (I2 – I1)
V =R 1
1 1
1 1
I1 I2
V 2
21 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.